If a solution(s) exists y=y so we can say:

x^2-3x=-2x+2 add 2x to both sides

x^2-x=2 subtract 2 from both sides

x^2-x-2=0 factor

(x-2)(x+1)=0

So x=-1 and 2, using y=-2x+2 we find:

y(-1)=4 and y(2)=-2

So the two solutions occur at the points:

(-1,4) and (2,-2)

4 0

4 years ago

Read 2 more answers

Find the square: (5x - 4y)2 (and please put the process of solving the equation)

satela [25.4K]

$(5x-4y)^2$
$(5x-4y)(5x-4y)$

Multiply each term in one bracket in turn by the terms in the other bracket

$(5x)(5x) = 25x^2$
$(5x)(-4y)=-20xy$
$(-4y)(5x)=-20xy$
$(-4y)(-4y)=16y^2$

Put them together we have
$25x^2-20xy-20xy-16y^2$
$25x^2-40xy-16y^2$

7 0

3 years ago

Decompose 549 two ways

RUDIKE [14]

Answer:

not sure if this is what you mean but you could do 500+49 or 510+39

4 0

3 years ago